Mechanics of Materials

Beam deflection and stress in members

Beam deflection is obtained by integrating the moment-curvature relation EI\,y'' = M(x) and applying the support conditions. A fixed end fixes both the deflection and the slope to zero, while a pin or roller fixes only the deflection. These conditions determine the two constants of integration. For a cantilever loaded by an end couple M, the bending moment is the same at every section, so the elastic curve is the parabola y = Mx^2/(2EI). The deflection at mid-span is therefore ML^2/(8EI), which is smaller than the tip value.

The moment-area method states that the change in slope between two points equals the area of the M/EI diagram between them. For a cantilever carrying a load at its free end, the tip slope is \theta = \frac{1}{EI}\int_0^L M\,dx, which is the area of the bending-moment diagram divided by EI.

Stiffness is the load divided by the corresponding displacement, and its reciprocal is compliance. For a cantilever under a tip load W, the tip deflection is WL^3/(3EI), so the effective stiffness is k = 3EI/L^3.

The unit-load method gives the displacement at a point as \delta = \int \frac{M\,\bar m}{EI}\,dx, where \bar m is the moment produced by a unit dummy load applied at the point and in the direction of the required displacement. For frames and overhangs, each member and each segment must be integrated separately, and the signs of M and \bar m must be taken consistently.

When the free end of an overhanging beam carries a mass, the stiffness k = F/\delta follows from the unit-load deflection, and the natural frequency of the single-degree-of-freedom system is \omega_n = \sqrt{k/m}.

When a support reaction is redundant, Castigliano's theorem gives it from \partial U/\partial R = 0 at the point where the displacement is prescribed. A pin that slides in a smooth slot has zero displacement in the direction normal to the slot, so the slot reaction is found by setting that displacement to zero.

Shear force and bending moment are related by dM/dx = V. The maximum bending moment therefore occurs where the shear force passes through zero. A concentrated tip load produces a bending moment that varies linearly and vanishes at the free end, whereas a uniformly distributed load produces a parabolic moment with zero slope at the free end. Moments due to loads in two perpendicular planes act about two different axes and must be drawn separately.

For a pin-roller beam with a linearly varying load, the reactions come from the three equilibrium equations. The shear is then integrated along the span, and the maximum moment is found at the section where the shear is zero.

A uniform distributed couple of intensity m per unit length on a simply supported beam of length L has a total couple mL. The reactions form an equal and opposite couple, so the internal bending moment is zero at every section.

For a prismatic bar with several axial loads, the internal force in each segment is found by summing the external loads on one side of a section. The axial stress \sigma = N/A is largest where |N| is largest.

For bending stress, \sigma = My/I, and the maximum stress occurs at the extreme fibre where y = c, giving \sigma_{max} = Mc/I. For a rectangular section of width b and depth h, I = bh^3/12.

For a rectangular section the shear stress is parabolic and reaches its maximum at the neutral axis, where \tau_{max} = 1.5\,V/A.

Self-weight deflection of a tapered member is found by integrating the local strain along its length. For a solid cone standing on its base, the compressive stress at distance x from the apex is wx/3, where w is the specific weight. Integrating this strain from the section to the base gives the deflection at mid-height as wH^2/(8E).

• Flexural rigidity — the product EI that governs bending deflection; the deflection scales inversely with it. e.g., doubling EI halves the tip deflection of a cantilever under a given load.
• Compliance — deflection per unit load, the reciprocal of stiffness. e.g., a cantilever with compliance L^3/(3EI) has stiffness 3EI/L^3.
• Extreme fibre — the point of the section farthest from the neutral axis, where the bending stress is largest.

Distortion energy theory (von Mises)

The distortion energy (von Mises) theory states that yielding begins when the distortion energy per unit volume reaches the value at yield in uniaxial tension. Distortion energy is the part of strain energy that changes the shape of an element without changing its volume. For an isotropic material it is U_d = \frac{1+\nu}{3E}\sigma_{vm}^2, so the von Mises stress is proportional to the square root of the distortional energy. The dilatational part, which depends only on the volume change, does not enter the criterion.

For plane stress with normal stresses \sigma_x, \sigma_y and shear \tau_{xy}, the equivalent stress is \sigma_e = \sqrt{\sigma_x^2 - \sigma_x\sigma_y + \sigma_y^2 + 3\tau_{xy}^2}. Yielding is predicted when \sigma_e equals the yield strength S_y. Substituting the stresses into this expression and solving for the load parameter gives the load at which yield starts.

In design, the permissible stress is the yield strength divided by the factor of safety, so the condition becomes \sigma_e \le S_y/n.

For pure shear, the von Mises criterion gives a shear yield stress k = \sigma_y/\sqrt{3}, which is about 0.577\,\sigma_y. The Tresca (maximum shear stress) criterion gives k = \sigma_y/2, so it is more conservative.

Failure theories differ in their predictions. Maximum normal stress theory predicts yielding when the largest principal stress reaches S_y. Maximum shear stress (Tresca) theory predicts yielding when (\sigma_1-\sigma_3)/2 reaches S_y/2. For a stress state with principal stresses \sigma_1 > \sigma_2 > \sigma_3, the two theories can therefore give opposite predictions when \sigma_1 is below S_y but \sigma_1-\sigma_3 exceeds S_y.

• Distortion energy — the strain energy associated with shape change at constant volume.
• Dilatational energy — the strain energy associated with volume change only.
• Equivalent (von Mises) stress — a single stress value that is compared with the uniaxial yield strength.

Thin-walled pressure vessel stresses

For a thin-walled closed cylinder of mean radius r and wall thickness t under internal pressure p, the hoop (circumferential) stress is \sigma_\theta = pr/t and the longitudinal (axial) stress is \sigma_z = pr/(2t). The radial stress is negligible compared with these, so the wall is treated as being in plane stress.

In the same vessel, the hoop strain is \varepsilon_\theta = (\sigma_\theta - \nu\sigma_z)/E. The thickness strain is \varepsilon_r = -\nu(\sigma_\theta+\sigma_z)/E. Because the hoop stress is twice the axial stress, the hoop strain is larger than the axial strain. When the allowable strain is checked, the larger strain component governs.

A welded seam of a cylinder is loaded by the same hoop and longitudinal stresses. If the weld normal lies at angle \theta to the axis, the normal stress on the weld is \sigma_n = \sigma_z\cos^2\theta + \sigma_\theta\sin^2\theta.

In an open-top vessel full of liquid, the pressure at depth y is p = \rho g y, so the hoop stress \sigma_\theta = \rho g y r/t increases linearly with depth and is zero at the brim. The axial stress in the wall is a load-transfer quantity that depends on how the vessel is supported, so it must be found from equilibrium of the wall with the liquid weight, not from the pressure alone.

• Hoop stress — the circumferential stress that resists the bursting tendency of the cylinder, equal to pr/t.
• Thin-walled — a wall whose thickness is small compared with the radius, so the stress is uniform through the thickness.
• Permissible pressure — the largest internal pressure that keeps the critical strain or stress below its allowable value.

Isotropic linear elastic relations

For a homogeneous, isotropic, linear elastic material, the normal strain in each direction is given by Hooke's law: \varepsilon_x = \frac{1}{E}[\sigma_x - \nu(\sigma_y+\sigma_z)], and the shear strain is \gamma = \tau/G, where G = E/[2(1+\nu)]. Shear stress and shear strain are coupled only through the shear modulus, so pure normal stresses produce no shear strain.

In plane stress, \sigma_z = 0, so the thickness strain is \varepsilon_z = -\frac{\nu}{E}(\sigma_x+\sigma_y). The shear stress does not contribute to this strain.

A body under hydrostatic pressure has equal normal stresses in all directions and therefore no shear stress. The volumetric strain for a uniaxial stress \sigma is \varepsilon_v = \varepsilon(1-2\nu). For ordinary metals with \nu \approx 0.3, this is positive, so a steel rod under tension gains volume.

For a beam, zero shear force means that dM/dx = 0, so the bending moment is constant and the curvature 1/\rho = M/EI is constant. The elastic curve is then a circular arc, which is straight only when M = 0.

For a linear elastic structure, the results of separate loadings can be added directly. The normal stress at a point under combined loading is the sum of the stress components produced by each loading, provided the loads do not change the geometry significantly.

The failure criteria for ductile and brittle materials are chosen differently. The maximum shear stress (Tresca) theory is used for ductile yielding. Brittle fracture is usually predicted by maximum normal stress or Mohr-type theories.

• Poisson's ratio — the negative ratio of lateral to axial strain; for common metals it lies between about 0.25 and 0.35, and it must be less than 0.5 for an isotropic material.
• Shear modulus — G = E/[2(1+\nu)] relates shear stress to shear strain.
• Superposition — the principle that stresses and strains from separate linear loadings add together.

Mohr's circle for plane stress

The Mohr's circle for plane stress has its centre at C = ((\sigma_x+\sigma_y)/2,\ 0) on the normal-stress axis, and its radius is R = \sqrt{[(\sigma_x-\sigma_y)/2]^2 + \tau_{xy}^2}. Because the centre always lies on the normal-stress axis, a valid circle is symmetric about that axis. Any circle whose centre is above or below the axis cannot represent a stress state.

The principal stresses are the points where the circle crosses the normal-stress axis, \sigma_{1,2} = C \pm R. A circle centred at the origin gives principal stresses +R and -R. A circle that passes through the origin with its centre at (R,0) gives principal stresses 2R and 0.

A uniform hydrostatic state, \sigma_x = \sigma_y = -p with \tau = 0, has zero radius, so its Mohr's circle collapses to a single point on the normal-stress axis at -p.

For a plane at angle \theta from the x-axis, the normal and shear stresses are \sigma_n = \frac{\sigma_x+\sigma_y}{2} + \frac{\sigma_x-\sigma_y}{2}\cos 2\theta + \tau_{xy}\sin 2\theta and \tau_n = -\frac{\sigma_x-\sigma_y}{2}\sin 2\theta + \tau_{xy}\cos 2\theta, with the sign of the shear depending on the convention. These expressions are used directly to find the stresses on a weld plane, and their ratio then gives the requested quantity.

• Centre of Mohr's circle — the average of the two normal stresses, always on the normal-stress axis.
• Degenerate circle — a circle of zero radius, which occurs when the two normal stresses are equal and the shear stress is zero.
• Principal plane — a plane on which the shear stress is zero.

Euler buckling of columns

The Euler critical load for a slender column is P_{cr} = \pi^2 EI/L_e^2, where L_e is the effective length. The critical load depends on the end conditions only through L_e, and it is inversely proportional to L_e^2.

The effective length depends on the end support conditions. A pin-pin column has L_e = L, a fixed-free column has L_e = 2L, a fixed-pinned column has L_e = L/\sqrt{2}, and a fixed-fixed column has L_e = L/2. The ratio of critical loads for two end conditions is therefore P_2/P_1 = (L_{e1}/L_{e2})^2.